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Cambridge Maths Academy
16. Integration of powers of the sine function (Wallis integral) 본문
16. Integration of powers of the sine function (Wallis integral)
Cambridge Maths Academy 2021. 12. 9. 08:29수학 모음 (Maths collection) 전체보기
Question. Let In=∫sinn(ax)dx(a) Show that In=−1ansinn−1(ax)cos(ax)+n−1nIn−2
Let Jn=∫π20sinnxdx
(b) Hence, or otherwise, show that Jn=n−1nJn−2 and hence that (i)n=2m(even):J2m=∫π20sin2mxdx=(2m)!22m(m!)2π2(ii)n=2m+1(odd):J2m+1=∫π20sin2m+1xdx=22m(m!)2(2m+1)!
Comment. We can generalise this result to: K2m=∫π2a0sin2m(ax)dx=(2m)!22m(m!)2π2aK2m+1=∫π2a0sin2m+1(ax)dx=22m(m!)2(2m+1)!1a
(Relevant: Edexcel 2017 Specification, FP2 Ch6 Reduction formulae)
Solution. (a) By integration by parts, u=sinn−1(ax)v′=sin(ax)u′=a(n−1)sinn−2(ax)cos(ax)v=−1acos(ax) we find In=∫sinn(ax)dx=−1asinn−1(ax)cos(ax)+(n−1)∫sinn−2(ax)cos2(ax)dx=−1asinn−1(ax)cos(ax)+(n−1)∫sinn−2(ax)(1−sin2(ax))dx=−1asinn−1(ax)cos(ax)+(n−1)∫sinn−2(ax)dx⏟In−2−(n−1)∫sinn(ax)dx⏟In=−1asinn−1(ax)cos(ax)+(n−1)In−2−(n−1)In⇒[1+(n−1)]In=−1asinn−1(ax)cos(ax)+(n−1)In−2⇒nIn=−1asinn−1(ax)cos(ax)+(n−1)In−2⇒In=−1ansinn−1(ax)cos(ax)+n−1nIn−2◻
(b) From (a) with a=1, Jn=∫π20sinnxdx=[−1nsinn−1xcosx]π20⏟=0+n−1nJn−2=n−1nJn−2◻
For the next part, we may approach it in three different ways.
Method 1.
We note the following results: J0=∫π20dx=π2J1=∫π20sinxdx=[−cosx]π20=−cos(π2)+cos(0)=1J2=∫π20sin2xdx=∫π20(1−cos2x2)dx=[12x−14sin2x]π20=π4(=12J0) (i) n=2m (even) J2m=2m−12mI2m−2=2m−12m⋅2m−32m−2J2m−4=2m−12m⋅2m−32m−2⋅2m−52m−4J2m−6=2m−12m⋅2m−32m−2⋅2m−52m−4⋯34⋅12J0⏟π2=(2m−1)(2m−3)(2m−5)⋯3⋅12m(2m−2)(2m−4)⋯4⋅2⋅π2=(2m)(2m−1)(2m−2)(2m−3)(2m−4)(2m−5)⋯3⋅2⋅1[2m(2m−2)(2m−4)⋯4⋅2]2π2=(2m)![2m⋅m(m−1)(m−2)⋯2⋅1]2π2=(2m)!22m(m!)2π2◻ (ii) n=2m+1 (odd) J2m+1=2m2m+1J2m−1=2m2m+1⋅2m−22m−1J2m−3=2m2m+1⋅2m−22m−1⋅2m−42m−3J2m−5=2m2m+1⋅2m−22m−1⋅2m−42m−3⋯21J1⏟=1=2m(2m−2)(2m−4)⋯4⋅2(2m+1)(2m−1)(2m−1)⋯3⋅1=[2m(2m−2)(2m−4)⋯4⋅2]2(2m+1)(2m)(2m−1)(2m−2)(2m−3)(2m−4)⋯3⋅2⋅1=[2m⋅m(m−1)(m−2)⋯2⋅1]2(2m+1)!=22m(m!)2(2m+1)!◻
Aside. As an exemplary application, we evaluate the first few integrals: J0=∫π20dx=π2J1=∫π20sinxdx=1J2=∫π20sin2xdx=π4J3=∫π20sin3xdx=23J4=∫π20sin4xdx=3π16J5=∫π20sin5xdx=815J6=∫π20sin6xdx=5π32J7=∫π20sin7xdx=1635J8=∫π20sin8xdx=35π256J9=∫π20sin9xdx=128315J10=∫π20sin10xdx=63π512J11=∫π20sin11xdx=256693
Method 2. Exponential form and binomial theorem:
We can also use the exponential definition of the sine function, sinx=eix−e−ix2i (i) n=2m (even) \begin{align} J_{2m} &= \int_0^{ \frac{ \pi }{ 2 } } \sin^{2m} x \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \left( \frac{ \textrm{e}^{ix} - \textrm{e}^{-ix} }{ 2i } \right)^{2m} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m}} \left( \textrm{e}^{ix} - \textrm{e}^{-ix} \right)^{2m} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m}} \left[ \sum_{r=0}^{2m} \binom{2m}{r} \textrm{e}^{i(2m-r)x} (-1)^r \textrm{e}^{-irx} \right] \, \textrm{d}x \qquad \textrm{by binomial theorem} \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} (-1)^r \underbrace{ \left( \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{2i(m-r)x} \, \textrm{d}x \right) }_{ I_r } \\ \end{align}
Here, we note that, for r \ne m, \begin{align} I_r &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{2i(m-r)x} \, \textrm{d}x \\ &= \left[ \frac1{2i(m-r)} \textrm{e}^{2i(m-r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= \frac{1}{2i(m-r)} \left( \textrm{e}^{i(m-r)\pi} - 1 \right) \\ \end{align} which gives (ignoring the subtlety with r=m for the next few lines as we're just trying to see the pattern for now) \begin{align} \Rightarrow \qquad J_{2m} &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} (-1)^r I_r \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r}{2i(m-r)} \left( \textrm{e}^{i(m-r)\pi} - 1 \right) \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r}{2i(m-r)} \Big( \cos[(m-r)\pi] - 1 \Big) \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r(-1)}{i(m-r)} \sin^2 \left( \frac{(m-r)\pi}{2} \right) \end{align} We observe a symmetry in the terms with r=s and r=2m-s, \begin{align} r&=s \;:& a_s&=\binom{2m}{s} \frac{(-1)^s(-1)}{i(m-s)} \sin^2 \left( \frac{(m-s)\pi}{2} \right) \\ \\ r&=2m-s \;:& a_{2m-s}&=\binom{2m}{2m-s} \frac{(-1)^{2m-s}(-1)}{i(s-m)} \sin^2 \left( \frac{(s-m)\pi}{2} \right) \end{align} and, since \binom{2m}{s}=\binom{2m}{2m-s}, we find \begin{align} a_s = - a_{2m-s} \end{align} This motivates to separate the sum into three parts, \begin{align} \sum_{r=0}^{2m} = \sum_{r=0}^{m-1} + \sum_{r=m}^{m} + \sum_{r=m+1}^{2m} \end{align} and in fact we should apply this separation before we perform the integral of I_r and include the case r=m as well: \begin{align} \Rightarrow \qquad J_{2m} &= \frac1{(2i)^{2m}} \left( \sum_{r=0}^{m-1} + \sum_{r=m}^{m} + \sum_{r=m+1}^{2m} \right) \binom{2m}{r} (-1)^r I_r \\ &= \frac{(-1)^m}{2^{2m}} \left[ \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r I_r + \binom{2m}{m} (-1)^m I_m + \sum_{r=m+1}^{2m} \binom{2m}{r} (-1)^r I_r \right] \end{align} For the last summation, let k = 2m-r \begin{align} \Rightarrow \qquad J_{2m} &= \frac{(-1)^m}{2^{2m}} \left[ \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r I_r + \binom{2m}{m} (-1)^m I_m + \sum_{k=0}^{m-1} \binom{2m}{2m-k} (-1)^{2m-k} I_{2m-k} \right] \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \Big( I_r + I_{2m-r} \Big) + \frac{(-1)^m}{2^{2m}} \binom{2m}{m} (-1)^m I_m \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \left( \frac{\textrm{e}^{i(m-r)\pi} - 1}{2i(m-r)} + \frac{\textrm{e}^{-i(m-r)\pi} - 1}{-2i(m-r)} \right) + \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \underbrace{ \left( \frac{ \cos[(m-r)\pi] - 1}{2i(m-r)} - \frac{\cos[(m-r)\pi] - 1}{2i(m-r)} \right) }_{ =0 } + \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{\pi}{2} \qquad \square \end{align}
(ii) n=2m+1 (odd) \begin{align} J_{2m+1} &= \int_0^{ \frac{ \pi }{ 2 } } \sin^{2m+1} x \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \left( \frac{ \textrm{e}^{ix} - \textrm{e}^{-ix} }{ 2i } \right)^{2m+1} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m+1}} \left( \textrm{e}^{ix} - \textrm{e}^{-ix} \right)^{2m+1} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m+1}} \left[ \sum_{r=0}^{2m+1} \binom{2m+1}{r} \textrm{e}^{i(2m+1-r)x} (-1)^r \textrm{e}^{-irx} \right] \, \textrm{d}x \qquad \textrm{by binomial theorem} \\ &= \frac1{(2i)^{2m+1}} \sum_{r=0}^{2m+1} \binom{2m+1}{r} (-1)^r \underbrace{ \left( \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{i(2m+1-2r)x} \, \textrm{d}x \right) }_{ I_r } \\ \end{align} We note that, \begin{align} I_r &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{i(2m+1-2r)x} \, \textrm{d}x \\ &= \left[ \frac1{i(2m+1-2r)} \textrm{e}^{i(2m+1-2r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= \frac1{i(2m+1-2r)} \left( \textrm{e}^{i(m-r)\pi + \frac{i\pi}{2} } - 1 \right) \\ &= \frac{ i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } \\ \\ I_{2m+1-r} &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{-i(2m+1-2r)x} \, \textrm{d}x \\ &= \left[ - \frac1{i(2m+1-2r)} \textrm{e}^{i(2m+1-2r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= - \frac1{i(2m+1-2r)} \left( \textrm{e}^{-i(m-r)\pi - \frac{i\pi}{2} } - 1 \right) \\ &= - \frac{ - i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } \\ &= \frac{ i \cos [(m-r)\pi] + 1 }{ i(2m+1-2r) } \\ \end{align} As with the even case above, we separate the summation into two parts, \begin{align} \sum_{r=0}^{2m+1} = \sum_{r=0}^{m} + \sum_{r=m+1}^{2m+1} \end{align} \begin{align} \Rightarrow \qquad J_{2m+1} &= \frac1{(2i)^{2m+1}} \left( \sum_{r=0}^{m} + \sum_{r=m+1}^{2m+1} \right) \binom{2m+1}{r} (-1)^r I_r \\ \end{align} For the second summation, let k = 2m+1-r and it gives \begin{align} \Rightarrow \qquad J_{2m+1} &= \frac{1}{(2i)^{2m+1}} \left[ \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r I_r + \sum_{k=0}^{m} \binom{2m+1}{2m+1-k} (-1)^{2m+1-k} I_{2m+1-k} \right] \\ &= \frac{1}{(2i)^{2m+1}} \left[ \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r I_r + \sum_{k=0}^{m} \binom{2m+1}{k} (-1)^{k-1} I_{2m+1-k} \right] \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r \Big( I_r - I_{2m+1-r} \Big) \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r \left( \frac{ i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } - \frac{ i \cos [(m-r)\pi] + 1 }{ i(2m+1-2r) } \right) \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^{r+1} \left( \frac{ 2 }{ i(2m+1-2r) } \right) \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m} \binom{2m+1}{r} \frac{ (-1)^{r} }{ (2m+1-2r) } \qquad \textrm{(Dirichlet beta function?)}\\ &= \frac{(-1)^m}{2^{2m}} \left( \frac{ 2^{2m+1} \Gamma \left( \frac12 - m \right) \Gamma(m+1) }{ \sqrt{\pi} (2m+1) } \right) \\ &= \frac{ 2(-1)^m m! }{ \sqrt{\pi} (2m+1) } \Gamma \left( \frac12 - m \right) \\ &= \frac{ 2(-1)^m m! }{ \sqrt{\pi} (2m+1) } \left[ (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi } \right] \\ &= \frac{ 2^{2m+1} (m!)^2 }{ (2m+1)! } \qquad \square \end{align} since \begin{align} \Gamma \left( -m + \frac12 \right) &= \frac{ \Gamma \left( -m +1 + \frac12 \right) }{ \left( -m + \frac12 \right) } \\ &= \frac{ \Gamma \left( -m + m + \frac12 \right) }{ \left( -m + \frac12 \right) \left( -m + 1 + \frac12 \right) \left( -m + 2 + \frac12 \right) \cdots \left( -m + (m - 1) + \frac12 \right) } \\ &= \frac{ \Gamma \left( \frac12 \right) }{ (-1)^m \frac{ (2m-1)(2m-3)(2m-5) \cdots 3 \cdot 1 }{ 2^m } } \\ &= \frac{ \sqrt{\pi} }{ (-1)^m \frac{ (2m)(2m-1)(2m-2)(2m-3)(2m-4) \cdots 1 }{ 2^m (2m)(2m-2)(2m-4) \cdots 2 } } \\ &= \frac{ \sqrt{\pi} }{ (-1)^m \frac{ (2m)! }{ 2^{2m} m! } } \\ &= (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi } \end{align}
Method 3. Integration by substitution and the Euler beta and gamma functions:
Use the substitution \begin{align} u &= \sin x \\ \textrm{d}u &= \cos x \, \textrm{d}x = \sqrt{ 1 - \sin^2x }\, \textrm{d}x = \sqrt{ 1 - u^2 }\, \textrm{d}x \end{align} and the integral reads \begin{align} \Rightarrow \qquad J_n &= \int_0^{\frac{\pi}{2}} \sin^nx \, \textrm{d}x = \int_0^1 u^n \left( 1 - u^2 \right)^{-\frac12} \, \textrm{d}u \end{align} We make the substitute: \begin{align} t &= u^2 \\ \textrm{d}t &= 2u \, \textrm{d}u = 2t^{\frac12} \, \textrm{d}u \end{align} which gives \begin{align} \Rightarrow \qquad J_n &= \int_0^1 t^{ \frac{n}{2} } \left( 1 - t \right)^{-\frac12} \, \frac{ \textrm{d}t }{ 2t^{\frac12} } \\ &= \frac12 \int_0^1 t^{ \frac{n}{2} - \frac12 } \left( 1 - t \right)^{-\frac12} \, \textrm{d}t \\ &= \frac12 B \left( \frac{n+1}{2}, \frac12 \right) \\ &= \frac12 \frac{ \Gamma\left( \frac{n}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{n}{2}+1 \right) } \end{align} where the Euler beta and gamma functions are defined by \begin{align} B(p,q) &= \int_0^1 t^{p-1} (1-t)^{q-1} \, \textrm{d}t \\ \Gamma(p) &= \int_0^\infty t^{p-1}\textrm{e}^{-t} \, \textrm{d}t \end{align} and \begin{align} B(p,q) = \frac{ \Gamma(p) \Gamma(q) }{ \Gamma(p+q) } \qquad \textrm{and} \qquad \Gamma(p+1) = p \Gamma(p) \end{align} Noting that \Gamma \left( \frac12 \right) = \sqrt{\pi} (see below), we find, for m \in \mathbb N_0, \begin{align} \Gamma \left( m + \frac12 \right) &= \left( m - \frac12 \right) \Gamma \left( m - \frac12 \right) \\ &= \underbrace{ \left( m - \frac12 \right) \left( m - \frac32 \right) \cdots \left( \frac12 \right) }_{n\;\textrm{terms}} \Gamma \left( \frac12 \right) \\ &= \frac{ (2m-1) (2m-3) \cdots 3 \cdot 1 }{ 2^m } \Gamma \left( \frac12 \right) \\ &= \frac{ (2m) (2m-1) (2m-2) (2m-3) (2m-4) \cdots 3 \cdot 2 \cdot 1 }{ 2^m (2m) (2m-2) (2m-4) \cdots 4 \cdot 2 } \underbrace{ \Gamma \left( \frac12 \right) }_{=\sqrt{\pi}} \\ &= \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \end{align} where we have used \begin{align} \left[ \Gamma \left( \frac12 \right) \right]^2 &= B\left( \frac12, \frac12 \right) \underbrace{ \Gamma(1) }_{0!=1} \\ &= \int_0^1 t^{-\frac12} (1-t)^{-\frac12} \, \textrm{d}t \\ &= \int_0^1 \frac1{ \sqrt{ t(1-t)} } \, \textrm{d}t \\ &= \int_0^1 \frac1{ \sqrt{ \frac14 - \left( t - \frac12 \right)^2 }} \, \textrm{d}t \end{align} and by the substitution \begin{align} t-\frac12 &= \frac12\sin v \\ \textrm{d}t &= \frac12\cos v\,\textrm{d}v \\ \end{align} which gives \begin{align} &\Rightarrow& \left[ \Gamma \left( \frac12 \right) \right]^2 &= \int_0^{\frac{\pi}{2}} \frac1{ \sqrt{ \frac14 \left( 1 - \sin^2v \right) }} \, \frac12\cos v\,\textrm{d}v = \pi \\ &\Rightarrow& \Gamma \left( \frac12 \right) &= \sqrt{ \pi } \end{align}
(i) n=2m (even) \begin{align} J_{2m}&= \frac12 \frac{ \Gamma\left( \frac{2m}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{2m}{2}+1 \right) } \\ &= \frac12 \frac{ \Gamma\left( m + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( m+1 \right) } \\ &= \frac12 \frac{ \left( \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \right) \sqrt{\pi} }{ m! } \\ &= \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{\pi}{2} \qquad \square \end{align}
(ii) n=2m+1 (odd) \begin{align} J_{2m+1}&= \frac12 \frac{ \Gamma\left( \frac{2m+1}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{2m+1}{2}+1 \right) } \\ &= \frac12 \frac{ \Gamma\left( m + 1 \right) \Gamma\left( \frac12 \right) }{ \Gamma \left( m + \frac32 \right) } \\ &= \frac12 \frac{ m! \sqrt{\pi} }{ \left( m + \frac12 \right) \Gamma \left( m + \frac12 \right) } \\ &= \frac12 \frac{ m! \sqrt{\pi} }{ \left( m + \frac12 \right) \left[ \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \right] } \\ &= \frac{ 2^{2m+1} (m!)^2 }{ (2m+1)! } \qquad \square \end{align}
A summary:
\begin{align} \int_0^{\frac{\pi}{2}}\sin^{2m}x\,\textrm{d}x=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2} \\ \\ \int_0^{\frac{\pi}{2}}\sin^{2m+1}x\,\textrm{d}x=\frac{2^{2m}(m!)^2}{(2m+1)!} \\ \\ \int_0^{\frac{\pi}{2}}\cos^{2m}x\,\textrm{d}x=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2} \\ \\ \int_0^{\frac{\pi}{2}}\cos^{2m+1}x\,\textrm{d}x=\frac{2^{2m}(m!)^2}{(2m+1)!} \end{align}
and \begin{align} \Gamma \left( \frac12 \right) &= \sqrt{ \pi } \\ \Gamma \left( m + \frac12 \right) &= \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi}, \quad m \in \mathbb Z_0^+ \\ \Gamma \left( -m + \frac12 \right) &= (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi }, \quad m \in \mathbb Z_0^+ \end{align}
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